1132 Cut Integer(20 分)

Cutting an integer means to cut a K digits lone integer Z into two integers of (K/2) digits long integers A and B. For example, after cutting Z = 167334, we have A = 167 and B = 334. It is interesting to see that Z can be devided by the product of A and B, as 167334 / (167 × 334) = 3. Given an integer Z, you are supposed to test if it is such an integer.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤20). Then N lines follow, each gives an integer Z (10 ≤ Z <2​31​​). It is guaranteed that the number of digits of Z is an even number.

Output Specification:

For each case, print a single line Yes if it is such a number, or No if not.

Sample Input:

3
167334
2333
12345678

Sample Output:

Yes
No
No

 

 

 

这题其实很简单,要注意在求余的时候a%b b不能是0

这题可以用stoi; 和to_string 正好是相反的意思

#include
#include
#include
#include
#include
#include
#include
using namespace std;

int main()
{
	int n;
	scanf("%d", &n);
	for (int i = 0;i < n;i++) {
		string s;
		cin >> s;
		string s1;
		string s2;
		int len = s.length();
		for (int j = 0;j < len/2;j++) s1.push_back(s[j]);
		for (int j = len/2;j < len;j++) s2.push_back(s[j]);
		int num = 0;
		int num1 = 0;
		int num2 = 0;
		for (int j = 0;j < s.size();j++) num = num * 10 + s[j] - '0';
		for (int j = 0;j < s1.size();j++)  num1 = num1 * 10 + s1[j] - '0';
		for (int j = 0;j < s2.size();j++) num2 = num2 * 10 + s2[j] - '0';
		//cout << num << " " << num1 << " " << num2 << endl;
		if (num1*num2!=0&&num % (num1*num2)==0) printf("Yes\n");
		else printf("No\n");

	}
	return 0;
}

 

 

#include
#include
#include
#include
#include
#include
#include
using namespace std;

int main()
{
	int T;
	scanf("%d", &T);
	while (T--) {
		int n;
		scanf("%d", &n);
		string s = to_string(n);
		int len = s.length();
		int s1 = stoi(s.substr(0, len / 2));
		int s2 = stoi(s.substr(len / 2));
		if (s1*s2 != 0 && n % (s1*s2) == 0) printf("Yes\n");
		else printf("No\n");
	}
	return 0;
}

 

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